Question

Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 2003, the average LOS for non-heart patients was 4.6 days. A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2008 of 3.8 days and a standard deviation of 1.2 days. Calculate a 95 percent confidence interval for the population mean LOS for non-heart patients in these hospitals in 2008. 41) ______

A) [3.24, 4.36] B) [3.27, 4.33] C) [3.34, 4.26] D) [3.38, 4.22]

Answer 1

Solution :

Given that,

= 3.8

s = 1.2

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,19 =2.093

Margin of error = E = t_/2,df * (s /n)

= 2.093 * (1.2/ 20) = 0.56

The 95% confidence interval estimate of the population mean is,

- E < < + E

3.8 - 0.56 < < 3.8 + 0.56

3.24 < < 4.36

correct option is A)(3.24, 4.36 )